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A buffer is a an aqueous solution formed from a weak conjugate acid-base pair that resists ph change upon the addition of another acid or base. after addition of H 3 O + equal concentrations of weak acid and its conjugate base after addition of OH - Add H 3 O + Add OH - CH 3 CO 2 H + OH - <==> CH 3 CO 2 - + H 2 O CH 3 CO 2 - + H 3 O + <==> CH 3 CO 2 H + H 2 O HA scavenges OH - while A - scavenges H +
Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (d) Na 2 HPO4/H3PO4 e)hcl/ch3coona
Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (d) Na 2 HPO4/H3PO4 A buffer requires a weak acid-conjugate base pair or the same combination generated in-situ. (a) KF is a weak acid and F - is its conjugate base. It s a buffer solution. (b) HBr is a strong acid, KBr is the salt of a strong base and strong acid. It can not be a buffer solution. (c) CO 3 2- is the conjugate base of HCO3 -. HCO 3 - is it conjugate acid. It is a buffer solution. (d) HPO 4 2- is a conjugate base of H 3 PO4. It is a conjugate acid--base pair and a buffer. (e) CH3COO - is conjugate base of CH3COOH. The HCl will generate the acid. It is a buffer.
The Buffer Equation (Henderson-Hasselbach) is key to understanding buffers. HA(aq) + H2O(l) K a = [H 3O + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A- ] [HA] H3O + (aq) + A - (aq) K a for acid HA Solve for [H + ] Take -log of both sides standard equilibrium for HA Move the minus sign into the log ph = pk a + log [A- ] [HA] ph = pk a + log [base] [acid]
If you understand weak-acid-conjugate base pairs Ka, then you can understand buffers! HA(aq) + H2O(l) H3O + (aq) + A - (aq) acid base conjugate conjugate acid base K a = [H 3O + ][A - ] [HA] pka = -log K a this information is found in a Ka table or it must be given. ph = pk a + log [A- ] [HA] ph = pk a + log [base] [acid] The A - and HA are concentrations of the weak acidconjugate base. It s a dilution/stoich problem.
Buffer Capacity and Buffer Range Buffer capacity is the amount of acid (base) that a buffer can neutralize before the buffer solution ph changes appreciably. Maximum capacity occurs when [HA] and [A - ] are large and approximately equal to each other. Acetic-acid/Acetate Buffer Add strong base and note ph to each buffer. Buffer range is the ph range over which a buffer effectively neutralizes added acids and bases. Practically speaking, the buffer range is +/- 1 units about the pk a of the acid.
We can also prepare alkaline buffer solutions by mixing a weak base with its conjugate acid just like we do with an acid-conjugate base buffer system. B + H2O <==> BH + + OH - K b = [BH+ ][OH - ] [B] [OH - ] = K b [B] [BH + ] -log [OH - ] = -log K b - log [B] [BH + ] -log [H + ] = -log K b + log [BH+ ] [B] poh = pk b + log [BH+ ] [B] Kb for base, B Solve for [OH - ] Take -log of both sides Move the minus sign into the log poh = pk b + log standard equilibrium for base B [conjugate acid] [base]
Show using a chemical equation or use LeChatlier s principle and explain why a solution of aqueous NH3 + NH4Cl constitutes a buffer solution. Over what ph range should this buffer work? NH3(aq) + H3O + <===> NH4 + (aq) + H2O(aq) NH3(aq) + H2O <===> NH4 + (aq) + OH - (aq)
Show using a chemical equation or use LeChatlier s principle and explain why a solution of aqueous NH3 + NH4Cl constitutes a buffer solution. Over what ph range should this buffer work? NH3(aq) + H2O <===> NH4 + (aq) + OH - (aq) K b = [NH 4 + ][OH - ] [NH3 ] = 1.8 X 10-5 pk b = -log(1.8 x 10-5 ) = 4.74, pka = 14.00-4.74 = 9.26 This ammonia/ammonium salt buffer solution would work well +/- 1 ph from 9.26 or from ~8.3-10.3! --Buffers work best when the concentrations of the acid and conjugate base are large and equal!
Buffers can be made in many equivalent ways---just make sure you get the weak acid-conjugate base pair! Weak acid + strong base Weak base + strong acid Weak acid + salt of acid Weak base + salt of base Salt of acid + strong acid Salt of base + strong base
We use the buffer equation and the 4-easy steps to prepare a buffer. 1. Choose a weak acid/base that has a pka as close as possible to the ph you wish to buffer ph. The conjugate base of the chosen weak acid is the salt needed for the other component. 2. Calculate the ratio of buffer component concentrations. Optimal buffering effect is realized when the ratio of A - to HA is 1:1 3. Determine the buffer concentration: [HA] + [A - ] 4. Dilute the solution and adjust the ph. ph = pk a + log [base] [acid]
An environmental chemist requires a carbonate buffer of ph 10.00. How many grams of solid Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO 3 to prepare the buffer? K a of HCO 3 - is 4.7 x 10-11.
An environmental chemist requires a carbonate buffer of ph 10.00. How many grams of Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO 3 to prepare the buffer? K a of HCO 3 - is 4.7 x 10-11. Identify the equilibrium and the acid-conjugate base pair and Ka HCO 3- (aq) + H 2 O(l) [conjugate base] ph = pk a + log [acid] [CO3 10 = 10.33 + log 2- ] [HCO 3 - ] CO 3 2- (aq) + H 3 O + (aq) pk ahco3 = - log(kahco3) = - log(4.7 x 10-11 ) = 10.33-0.33 = log [CO3 2- ] [HCO3 - ] 10-0.33 = CO3 2-0.20 [CO3 2- ] = 0.20 X 10-0.33 = 0.0935 M
An environmental chemist needs a carbonate buffer of ph 10.00. How many grams of Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO 3 to prepare the buffer? K a of HCO 3 - is 4.7 x 10-11. [CO3 2- ] = 0.20 X 10-0.33 = 0.0935 M We need 1.5L of 0.0935M CO3 2-. It s now a molarity problem. mol CO 3 2- = (1.5 L)(0.0935 mol CO3 2- /L) = 0.1402 moles CO 3 2- mol Na2CO3 = 0.140 mol CO3 2-1 mol Na2CO3 X = 0.14 mol Na2CO 2-1 mol of CO 2-3 3 0.14 mole Na2CO3 X 105.99 g mol Na2CO3 = 15 g Na 2 CO 3
You wish to prepare a 1.0L buffer with a ph of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component? Acid Conjugate Base Ka pka HSO4 - SO4-1.2 x 10-2 1.92 CH3CO2H CH3CO2-1.8 x 10-5 4.74 HCO3 - CO3-2 4.8 x 10-11 10.32
You wish to prepare a 1.0L buffer with a ph of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component? Acid Conjugate Base Ka pka HSO4 - SO4-1.2 x 10-2 1.92 CH3CO2H CH3CO2-1.8 x 10-5 4.74 HCO3 - CO3-2 4.8 x 10-11 10.32 1. Choose pair with pka +/- 1 within desired ph. Acetic Acid-acetate is the only choice that fits this criteria. 2. Use the buffer equation ph = pk a + log [HCOO- ]initial [HCOOH]initial = 3.90 = 4.74 + log [HCOO- ]initial [HCOOH]initial 3.90-4.74 = log [HCOO- ]initial [HCOOH]initial 10-0.84 = [HCOO- ]initial [HCOOH]initial =.144
You wish to prepare a 1.0L buffer with a ph of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component? 3.90-4.74 = log [HCOO- ]initial [HCOOH]initial 10-0.84 = [HCOO- ]initial [HCOOH]initial =.144 [HCOO - ] =.144[HCOOH] 0.5M = total concentation = [HCOOH] + [HCOO - ] 0.5M = total concentation = [HCOOH] + 0.144[HCOOH] 0.5M = total concentation = 1.144[HCOOH] [HCOOH] =.5M/1.144 = 0.437M [HCOO-] =.5-0.437 = 0.063
What is the ph of a solution containing 0.30 M HCOOH (pk a = 3.77) and 0.52 M HCOOK (potassium formate)?
Starting with 0.5M formic acid and 0.50M NaOH stock solutions, how would you prepare 100. ml of 0.10M formic acid-formate buffer, ph = 4.25? The pka of formic acid is 3.70. KEY: 0.10M buffer = [A - ] + [HA]; [x] + [0.1 -x] ph = pk a + log [HCOO- ]initial [HCOOH]initial
Starting with 0.5M formic acid and 0.50M NaOH stock solutions, how would you prepare 100. ml of 0.10M formic acid-formate buffer, ph = 4.25? The pka of formic acid is 3.70. KEY: 0.10M buffer = [A - ] + [HA]; [x] + [0.1 -x] ph = pk a + log [HCOO- ]initial [HCOOH]initial x = 3.548 (0.1 -x) [x] 4.25 = 3.70 + log x =.3548-3.548x [.1 -x] [x] x =.3548-3.548x 4.25-3.70 = log [.1 -x] x + 3.548x =.3548 [x] 0.55 = log 4.548x =.3548 [.1 -x] x =.3548 10 0.55 [x] 4.548 = [.1 -x] [A-] = [formate] = x = 0.0780M [x] 3.548 = [HA] = [acid] = 0.1 - x = 0.1-0.0780 [.1 -x] = 0.022M
Starting with 0.5M formic acid and 0.50M NaOH, how would you prepare 100. ml of 0.10M formic acid-formate buffer, ph = 4.25? The pka of formic acid is 3.70. We need 100. ml of buffer with these concentrations: x = 0.0780M = [formate] [formic acid] = 0.022M Mol HCOOH = 0.022M HCOOH X 0.100 L = 0.0022M HCOOH This must come from the 0.5M HCOOH stock solution ml stock = 0.0022M HCOOH X 1000mL/0.5M HCOOH = 4.4 ml Moles HCOO - = 0.0780M HCOO - X 0.100 L = 0.0078M HCOO We get this from 1:1 neutralization with 0.50M NaOH COOH + NaOH => COO - + H2O 0.0078M NaOH must come from 0.50M NaOH ml NaOH stock = 0.0078M HCOOH X 1000mL/0.5M NaOH = 15.6 ml 0.0078M HCOO must come from acid stock ml stock = 0.0078M HCOOH X 1000mL/0.5M HCOOH = 15.6 ml
What is the ph of a solution containing 0.30 M HCOOH (pk a = 3.77) and 0.52 M HCOOK (potassium formate)? Initial (M) Change (M) Equilibrium (M) HCOOH (aq) 0.30 0.00 -x +x 0.30 - x H + (aq) + HCOO - (aq) x 0.52 +x 0.52 + x pka = 3.77 = - log (Ka) x (0.52 + x) K a = [H+ ] [HCOO - ] = 10-3.77 = Ka = 1.70 X 10-4 [HCOOH] 0.30 - x Can we simplify--yes! K a x 100 = 1.70 x 10-4 x 100 = 1.7 x 10-2 << [.30] and [.52] K a = 1.70 x 10-4 = x (0.52) 0.30 x = 0.30 1.7 X 10-4 = 9.80767 X 10-5 (0.52) ph = -log(h + ) = -log(9.80767 X 10-5 ) = 4.01
A) Calculate the ph of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. The pkb of ammonia is 4.75. NH 3 + H2O NH 4 + (aq) + H2O pk w = pka + pkb = 14 NH 4 + (aq) + OH - (aq) H + (aq) + NH 3 (aq) pk b = 4.75 pk a =? pka = 14.00 - pkb pka = 14.00-4.75 = 9.25 ph = pk a + log [conjugate base] [acid] ph = pk a + log [NH 3 ] [NH 4+ ] ph = 9.25 + log [0.30] [0.36] = 9.17